1) A block of mass 1.05 kg is placed on top of a second block of mass 7.84 kg. T

Question

1) A block of mass 1.05 kg is placed on top of a
second block of mass 7.84 kg. The coefficient
of kinetic friction between the 7.84 kg block
and the surface is 0.19. A horizontal force F
is applied to the 7.84 kg block.Calculate the magnitude of the force neces-
sary to pull both blocks to the right with an
acceleration of 4.02 m/s2. Assume no slipping
between the two blocks. The acceleration of
gravity is 9.8 m/s2 .
Answer in units of N

2) Find the minimum coefficient of static friction
µt between the blocks such that the 1.05 kg
block does not slip under an acceleration of
4.02 m/s2.

Explanation / Answer

F = 0.19*8.89*9.8 + 8.89*4.02 = 52.29 N Find the minimum coefficient of static friction µt between the blocks such that the 1.05 kg block does not slip under an acceleration of 4.02 m/s2 = 4.02/9.8 = 0.410

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